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    Solve cosec22x - cot2x = 1 between 0 and 180 degrees.

    I managed to factorise this to cot2x(cot2x -1)

    so then cot2x = 0 and cot2x = 1

    However, for cot2x = 0, I rewrote this as 1/tan2x = 0 but apparently, you have to rewrite it as cos2x/sin2x = 0 instead.

    Can someone explain why you can't use 1/tan2x = 0 for this?
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    (Original post by jessyjellytot14)
    Solve cosec22x - cot2x = 1 between 0 and 180 degrees.

    I managed to factorise this to cot2x(cot2x -1)

    so then cot2x = 0 and cot2x = 1

    However, for cot2x = 0, I rewrote this as 1/tan2x = 0 but apparently, you have to rewrite it as cos2x/sin2x = 0 instead.

    Can someone explain why you can't use 1/tan2x = 0 for this?
    Rare pitfall that is never really explained at A-Level. It's because \cot(x) \not\equiv \frac{1}{\tan(x)} so you cannot use tan directly, just turn it into sine and cosine instead.
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    (Original post by jessyjellytot14)
    Solve cosec22x - cot2x = 1 between 0 and 180 degrees.

    I managed to factorise this to cot2x(cot2x -1)

    so then cot2x = 0 and cot2x = 1

    However, for cot2x = 0, I rewrote this as 1/tan2x = 0 but apparently, you have to rewrite it as cos2x/sin2x = 0 instead.

    Can someone explain why you can't use 1/tan2x = 0 for this?
    You can't divide 1 by a value and get 0, but if it's \frac{cos(2x)}{sin(2x)} = 0 it follows that cos(2x)=0 , you could however use the values where tan x is undefined, as tanx approaches infinity for  x = \frac{\pi}{2} +2k \pi and \lim_{x\to\infty}\frac{1}{x} =0
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    The "identity" cotx = 1/tanx doesn't always hold. Cotx is defined to be cosx/sinx (where x is not an integer multiple of pi of course)
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    (Original post by jessyjellytot14)
    Solve cosec22x - cot2x = 1 between 0 and 180 degrees.

    I managed to factorise this to cot2x(cot2x -1) *
    this does not follow from the first line **
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    (Original post by the bear)
    this does not follow from the first line **
    It does but she didn't show it. Just move the 1 to the LHS and use identity \csc^2(2x)-1 \equiv \cot^2(2x) and factorise.
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    cos x isn't 1 over tan x always
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    (Original post by jessyjellytot14)
    Solve cosec22x - cot2x = 1 between 0 and 180 degrees.

    I managed to factorise this to cot2x(cot2x -1)

    so then cot2x = 0 and cot2x = 1

    However, for cot2x = 0, I rewrote this as 1/tan2x = 0 but apparently, you have to rewrite it as cos2x/sin2x = 0 instead.

    Can someone explain why you can't use 1/tan2x = 0 for this?
    In addition what was said above, you know that \cot(x) \equiv \frac{\cos(x)}{\sin(x)} which is fine, but remember that to get the \tan(x) you would need to divide top and bottom by cosine. The problem with this is that cosine can be 0, which is always an issue when it comes to division. To overcome this, we must strictly say that \cos(x) \not= 0 and only then will \cot(x) \equiv \frac{1}{\tan(x)}.

    Of course, however, you would be losing some solutions due to the restriction and \cot(x) \not= 0 as a result of our new 'identity' because \frac{1}{X} \not= 0, \forall X
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    (Original post by the bear)
    this does not follow from the first line **
    Yeah, I skipped a few steps to save typing it all up but I used the identity 1 + cot22x = cosec22x to remove the cosec22x and then the 1s on each side of the equation cancelled each other out. And then I factorised from there.
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    (Original post by jessyjellytot14)
    Yeah, I skipped a few steps to save typing it all up but I used the identity 1 + cot22x = cosec22x to remove the cosec22x and then the 1s on each side of the equation cancelled each other out. And then I factorised from there.
    OK i will let you off on this occasion *
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    (Original post by RDKGames)
    In addition what was said above, you know that \cot(x) \equiv \frac{\cos(x)}{\sin(x)} which is fine, but remember that to get the \tan(x) you would need to divide top and bottom by cosine. The problem with this is that cosine can be 0, which is always an issue when it comes to division. To overcome this, we must strictly say that \cos(x) \not= 0 and only then will \cot(x) \equiv \frac{1}{\tan(x)}.

    Of course, however, you would be losing some solutions due to the restriction and \cot(x) \not= 0 as a result of our new 'identity' because \frac{1}{X} \not= 0, \forall X
    Thank you, this is really helpful The textbook doesn't mention this.
 
 
 
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